Consider the spectral line resulting from the transition $n = 2 \rightarrow n = 1$ in the atoms and ions given below. The shortest wavelength is produced by:

The gravitational attraction between an electron and a proton in a hydrogen atom is weaker than the Coulomb attraction by a factor of about $10^{-40}$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction.

In a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is $r_n$. The graph between $\log \left( \frac{r_n}{r_1} \right)$ and $\log n$ will be:

The electron of a hydrogen atom makes a transition from the $(n + 1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to:

The inverse square law in electrostatics is $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}{r^2}}}$ for the force between an electron and a proton. The $\frac{1}{r^2}$ dependence of $|\vec F|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$,the force would be modified to $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\lambda }{r}} \right)\left( {{e^{ - \lambda r}}} \right)$ where $\lambda = \frac{{{m_p}c}}{\hbar }$ and $\hbar = \frac{h}{{2\pi }}$. Estimate the change in the ground state energy of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron.

Consider an electron in a hydrogen atom,revolving in its second excited state (having radius $4.65 \, \mathring{A}$). The de-Broglie wavelength of this electron is .... $\mathring{A}$.